A charge q is enclosed by a Gaussian spherical surface of radius R. If the radius is reduced to half of R, the outward electric flux will be
Answer & explanation
Correct answer: option 2
The correct answer is Option (2) → the same as its initial value
By Gauss's law the total electric flux through any closed surface depends only on the charge enclosed.
\( \displaystyle \Phi = \frac{q_{\mathrm{enc}}}{\varepsilon_0} \)
The smaller sphere still encloses the charge \( q \), so the flux is unchanged. Halving \( R \) makes the field on the surface four times larger (\( E \propto 1/R^{2} \)), but the area becomes one-fourth (\( A \propto R^{2} \)); the product \( EA \) stays the same.
NCERT (2026-27):
Gauss’s Law – “… (i) Gauss’s law is true for any closed surface, no matter what its shape or size. (ii) The term q on the right side of Gauss’s law, Eq. (1.31), includes the sum of all charges enclosed by the surface.”